In how many ways can the digits of 45,520 be arranged to form a 5-digit number? (Remember, numbers cannot begin with 0.
Is it 96?
4 choices for the 1st digit
4 choices for the 2nd digit
3 choices for the 3rd digit
2 choices for the 4th digit
However because there are 2 5's the number will be halved because of division by 2!
4x4x3x2/2 = 4x4x3 = 48 different 5 digit numbers
hmm, I have one shot and I'm trying to figure this one out. Rom, you say to divide by 2! but why do you divide by 2! when you don't use the second 5 at all?
Cool! You DO use the second '5' ...but it is idistiguishable from the other....so the numbers would look the same....so you must divide by two....
to expand on this a bit permutations of the same digits are indistinguishable.
5,5 is just 5,5, same with 5,5,5, or 5,5,5,5. They all have only a single permutation.
Whereas n different digits would have n! permutations.
When you go through and take the product of the possibilities like I did it assumes all permutations
are distinguishable. That's why you then have to divide the result by n!, in this case 2! as there are 2 5's.
If there were 3 5's I'd divide by 3! etc.
so 48 is the answer?